Solving Complex Number Point Set: |z|=3|z-1|

[elcotufa]

Homework Statement

Describe the set of points z in the complex plane that satisfies each of the following.

$$|z|=3|z-1|$$





$$x^2+y^2= 3[(x-1)^2+y^2]$$
$$x^2+y^2= 3[x^2-2x+1+y^2]$$
$$x^2+y^2= 3x^2-6x+3+3y^2$$
$$-2y^2= 2x^2-6x+3$$
$$-y^2= x^2-3x+\frac32$$
$$-\frac32-y^2=x(x-3)$$


then

$$x=-y^2-\frac32$$
or
$$x=-y^2+\frac32$$



Ta said it was wrong but I don't know why?

Does it has to be expressed with Y=?

Input appreciated

 

[jezznar]

i think what you need to do is to subtract the whole function to 1 not a single factor to 1.

i mean:

[x^2 + y^2 ] will be your [ z ] function here
not [ x ] = [ z ]

 

[Dick]

If x*(x-3)=0 then you can conclude x=0 or x-3=0. If x*(x-3)=B not equal to zero that DOES NOT MEAN x=B or x-3=B. Do you see why?

 

[HallsofIvy]

How about doing it geometrically? |z| is the distance from the complex number z= x+ iy, representing the point (x, y) in the plane to (0,0) and |z-3| is the distance from the complex number x+ iy to the number 3+ i0 or from (x,y) to (3, 0). What is the locus of points equidistant from (0, 0) and (3, 0)?

 

[Dick]

How about doing it geometrically? |z| is the distance from the complex number z= x+ iy, representing the point (x, y) in the plane to (0,0) and |z-3| is the distance from the complex number x+ iy to the number 3+ i0 or from (x,y) to (3, 0). What is the locus of points equidistant from (0, 0) and (3, 0)?

But (x,y) isn't equidistant from (0,0) and (1,0) (not (3,0)). The distance from (0,0) is three times the distance from (1,0). If the OP takes the fifth line of the attempt at solution and completes the square, it will be clear it's a circle.